Advertisements
Advertisements
प्रश्न
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
पर्याय
zero
`-(0.45 hati + 0.6 hatj)`
`-(0.9 hati + 1.2 hatj)`
`-5 (hati + hatj)`
Advertisements
उत्तर
`-(0.9 hati + 1.2 hatj)`
Explanation:
Given, `u = (3hati + 4hatj)` m/s
And `v = - (3hati + 4hatj)` m/s
Mass of the ball = 150 g = 0.15 kg
Δp = Change in momentum
= Final momentum – Initial momentum
= `mv - mu`
= `m(v - u) = (0.15) [- (3hati + 4hatj) - (3hati + 4hatj)]`
= `(0.15) xx [ - 6hati - 8hatj]`
= `- [0.15 xx 6hati + 0.15 xx 8hatj]`
= `- [0.9 hati + 1.20 hatj]`
Hence, Δp = `-[0.9 hati + 1.2 hatj]`
APPEARS IN
संबंधित प्रश्न
A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
A car accelerates on a horizontal road due to the force exerted by.
A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude.
A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will.
car moving at 40 km/hr is to be stopped by applying brakes in the next 4 m. If the car weighs 2000 kg, what average force must be applied to stop it?
Find the reading of the spring balance shown in the following figure. The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.

A constant force F = m2g/2 is applied on the block of mass m1 as shown in the following figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.

Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.

A tennis ball and a cricket ball , both are stationary. To start motion in them .
Two bodies A and B of same mass are moving with velocities v and 2v, respectively. Compare their (i) inertia and (ii) momentum.
Write the mathematical form of Newton's second law of motion. State the conditions if any.
How can Newton's first law of motion be obtained from the second law of motion?
A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : Change in momentum of the body.
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.
Multiple Choice Question. Select the correct option.
The impulse of a body is equal to:
Name the physical quantity which equals the rate of change of linear momentum.
A ball is thrown vertically downward with an initial velocity of 10 m/s. What is its speed 1 s later and 2 s later?
A metre scale is moving with uniform velocity. This implies ______.
A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms−1, q = 4 ms−2 and r = 5 ms−3. The force acting on the body at t = 2 seconds is ______.
